What is the extraneous solution to these equations? $\dfrac{x^2 - 2}{x - 3} = \dfrac{11x - 12}{x - 3}$
Multiply both sides by $x - 3$ $ \dfrac{x^2 - 2}{x - 3} (x - 3) = \dfrac{11x - 12}{x - 3} (x - 3)$ $ x^2 - 2 = 11x - 12$ Subtract $11x - 12$ from both sides: $ x^2 - 2 - (11x - 12) = 11x - 12 - (11x - 12)$ $ x^2 - 2 - 11x + 12 = 0$ $ x^2 + 10 - 11x = 0$ Factor the expression: $ (x - 10)(x - 1) = 0$ Therefore $x = 10$ or $x = 1$ The original expression is defined at $x = 10$ and $x = 1$, so there are no extraneous solutions.